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Question : 9 of 160
Marks:
+1,
-0
Solution:
‌‌ We have, ‌‌| (cos‌a+isin‌a)6 |
| (sin‌b+i‌cos‌b)8 |
‌=‌| (cos‌a+isin‌a)6 |
| (−i2sin‌b+i‌cos‌b)8 |
‌‌[∵1=−i2]‌=‌| (cos‌a+isin‌a)6 |
| i8(cos‌b−isin‌b)8 |
‌=‌| (cos‌a+isin‌a)6 |
| (cos‌b−isin‌b)8 |
‌=‌| cos‌6‌a+isin‌6a |
| cos‌8‌b−isin‌8b |
‌=‌| (cos‌6‌a+isin‌6a)(cos‌8‌b+isin‌8b) |
| (cos‌8‌b−isin‌8b)(cos‌8‌b+isin‌8b) |
‌∴‌ Required real part ‌‌=‌| cos‌6‌a‌cos8b−sin‌86asin‌8b |
| cos28b+sin‌28b |
‌=cos(6a+8b)
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